#### The diffraction of a crystal of Barium with X-ray of wavelength 2.29 $\dpi{100} \small \AA$ gives a first order reflection at $\dpi{100} \small 27^{\circ}8'$ . What is the distance (in $\AA$) between the diffracting planes ?Option: 1 1.46Option: 2 1.5Option: 3 2.51Option: 4 5.46

Bragg’s Equation

This equation gives a simple relationship between the wavelength Of X-rays and the distance between the planes in the crystal and the angle Of reflection. This equation can be written as:

$\\\mathrm{n} \lambda=2 \mathrm{d} \sin \theta\\\\ \mathrm{Here} \\\mathrm{n\:=\:Order\:of\:reflection;\:in\:general\:it\:is\:taken\:as\:1.}\\\\\mathrm{\lambda\: =\:Wavelength\:of\:X-rays}\\\\\mathrm{d\: =\:Distance\:between\:two\:layers\:of\:the\:crystals}\\\\\mathrm{\theta\: =\: Angle\:of\:incident\:light}$

As for a given set of lattice planes the value of 'd' is fixed so the possibility of getting maximum reflection depends only on θ. If we increase θ gradually a number of positions will be observed at which there will be maximum reflection.

Now,

$n\lambda=2d\sin\theta$

$n=1,\lambda=2.29\AA ,\theta=27^{\circ}8'$

So,

$d=0.5\times1\times10^{-8}/\sin(\theta=27^{\circ}8')=2.51\AA$